Help with taking the derivative of any of these inverse tangent functions?

PS I KNOW THE FORMULAS I'M JUST NEW TO THIS AND STILL DON'T QUITE UNDERSTAND IT ALL (Headbang)

1. g(x) = 3arccos(x/2)

I got:

-3/[2*sqrt(4-x^2)]

The answer is:

As above without the '2'

2. f(x) = arctan (x/a)

I got:

[1/a] / [1 + x^2/a^2]

The answer is:

a / [a^2+x^2]

3. g(x) = arcsin (3x) / x

NO IDEA

The answer is:

[3x - sqrt(1-9x^2)*arcsin(3x)] / [x^2*sqrt(1-9x^2)]

4. h(t) = sin[arccos(t)]

The answer is:

-t/sqrt(1-t^2)

5. y = x*arccos(x) - sqrt(1-x^2)

The answer is:

arccos(x)

6. y = (1/2) * [ (1/2)*ln{x+1/X-1} + arctan(x) ]

The answer is:

1/[1-x^4]